Question

The number of integral solutions of $\frac{x+1}{x^2+2}>\frac{1}{4}$, is

• A. $1$
• B. $2$
• C. $5$
• D. None of these

Answer 1

The correct option is C $5$

$\frac{x+1}{x^2+2}>\frac{1}{4}$

$\Rightarrow 4(x+1)>x^2+2$

$=x^2-4x+2-4<0$

$=x^2-4x+4-6<0$

$\Rightarrow {(x-2)}^2-{\left(\sqrt{6}\right)}^2<0$

$(x-2-\sqrt{6})(x-2+\sqrt{6})<0$

$(x-(2+\sqrt{6}\left)\right)$ $(x-(-\sqrt{6}+2\left)\right)<0$

$x\in (2-\sqrt{6},2+\sqrt{6})$

$\Rightarrow x\in (-0.4,4.4)$

$\Rightarrow x=0,1,2,3,4$

$\therefore x$ takes on $5$ integer values.

Hence, the answer is $5.$